Sulfur trioxide conditioning system control algorithm

ABSTRACT

Optimizing fly ash resistivity by controlling concentration of sulfur trioxide (SO 3 ) in flue gas by the use of an algorithm.

RELATED APPLICATION

This application is related to provisional application 60/338,152, filed Dec. 6, 2001, the contents of which are herein incorporated by reference.

FIELD OF THE INVENTION

The herein disclosed invention finds applicability in producing optimum fly ash resistivity in flue gas.

BACKGROUND OF THE INVENTION

Many utilities now burn a variety of coals at their fossil fuel plants. This practice is growing for several reasons, including (1) the need to lower SO₂ emissions by burning low-sulfur coals and (2) the need to reduce fuel costs to enhance their competitive position. Frequently, these coal changes have adverse affects on ESPs (ESP=electrostatic precipitator). Low-sulfur coals produce high resistivity ash that is difficult to collect in an electrostatic precipitator (the technology most commonly used to control particulate emissions from coal-fired power plants). Inexpensive coals are frequently variable in their properties and sometimes high in ash or low in sulfur. Conditioning the ash with SO₃ before the ash enters a precipitator, can lower ash resistivity and improve ESP performance. In fact, this well established technology is used at several hundred plants both here and abroad to control fly ash resistivity in low-sulfur or variable-sulfur coals. While commercial conditioning systems are relatively reliable, the controls for these are not sophisticated, and this lack of sophistication can result in non-optimum ESP performance, and sometimes excess SO₃ addition rates (and emissions).

The inventors have developed a correlation between certain ESP electrical operation parameters and fly ash resistivity. In particular, the inventors have found it to be possible to monitor the current density in an ESP electrical section, and using this number, estimate the resistivity of the fly ash in that electrical section. Using these correlations makes it possible to determine if the resistivity in this section is at an optimum level or not. Further, the inventors have also participated in the development of correlations between fly ash resistivity and the flue gas SO₃ concentration needed to produce optimum fly ash resistivity. These correlations can be combined (as described below) to produce a superior SO₃ conditioning system control algorithm.

Current SO₃ conditioning systems use a preset rate of SO₃ addition that is only adjusted for unit load. The invention, described herein, uses a unique combination of calculations to provide a rate of addition that is based on actual ESP operating data. These data, which can easily be obtained from modem ESP controls, are both real time and continuous. Hence, the new control algorithm is capable of producing an optimum rate of SO₃ addition when coal and ash properties are changing.

The primary application will be at utility plants that use SO₃ conditioning to improve ESP performance. These plants are located both here and abroad. In addition, SO₃ is used in some industrial applications, and the new SO₃ control algorithm could be used at those plants as well.

SUMMARY OF INVENTION

The herein disclosed invention is directed to a process for treating fly ash found in flue gas to produce effective fly ash electrical resistivity comprising employing an algorithm to determine the optimum amount of sulfur trioxide (SO₃) to be added to the flue gas. The sulfur trioxide can result from the burning of coal, or the sulfur trioxide can result from the burning of coal plus the extrinsic addition of sulfur trioxide. The process of this invention involves an algorithm which takes into account 1) flue gas SO₃ concentration, 2) initial fly ash resistivity, 3) electrostatic precipitator (ESP) current densities, 4) flue gas temperature and moisture and 5) fly ash composition.

Also, embraced by this invention is a process for treating fly ash found in flue gas to produce effective fly ash resistivity comprising the following steps:

Step 1. Obtain the proximate ultimate analyses of coal being burned in boiler and ash mineral analysis for this coal;

Step 2. Determine the average temperature of flue gas entering the electrostatic precipitator (ESP);

Step 3. Estimate SO₃ background level in the flue gas using correlation relating flue gas SO₃ to coal type and coal sulfur content.

Step 4. Calculate the base ash resistivity using empirical equations relating ash resistivity to ash composition, flue gas moisture and flue gas temperature.

Step 5. Use a correlation relating the base fly ash resistivity and flue gas SO₃ concentration to determine the flue gas SO₃ concentration needed to produce the optimum fly ash resistivity.

Step 6. Subtract the background SO₃ concentration from the needed SO₃ concentration from the needed SO₃ that must be added to the flue gas to produce the optimum fly ash resistivity and

Step 7. Send rate of addition signal to the controls that operate the SO₃ conditioning system.

Another method encompassed by the invention involves determining a most effective injection rate for SO₃ into flue gas comprising the following steps:

Step 1. Obtain the proximate and ultimate analysis of the coal being burned in the boiler and the ash mineral analysis for the coal,

Step 2. Determine the average temperature of the flue gas entering the ESP from plant instrumentation,

Step 3. Estimate SO₃ background level in the flue gas using correlation relating flue gas SO₃ to coal type and coal sulfur content,

Step 4. The secondary current applied to the electrostatic precipitator is obtained from the controls for each transformer-rectifier set that is powering the precipitators,

Step 5. Determine effective fly ash resistivity level in the ESP using a correlation that relates fly ash resistivity to ESP current density for each electrical field, average the results to produce an effective resistivity for the ESP.

Step 6. a. If indicated ash resistivity is equal to or less than optimum resistivity, decrease rate of injection by x percent where x is between 5 and 25, or

-   -   b. if indicated ash resistivity is greater than optimum         resistivity, increase rate of injection by x percent where x is         between 5 and 25.

Step 7. Repeat Step 6 until indicated fly ash resistivity passes through optimum resistivity point and then set rate of injection at a point in the range bounded by the levels calculated in the last two interactions, and then

Step 8. Every y minutes, where y is number between 5 and 30, restart the process beginning at Step 2.

A still further method involves a method for determining a most effective injection rate for SO₃ into flue gas comprising the following steps,

Step 1. Obtain the proximate and ultimate analysis of the coal being burned in the boiler and the ash mineral analysis for the coal,

Step 2. Determine the average temperature of the flue gas entering the ESP from plant instrumentation,

Step 3. Estimate SO₃ background level in the flue gas using correlation relating flue gas SO₃ to coal type and coal sulfur content,

Step 4. The secondary current applied to the electrostatic precipitator is obtained from the controls for each transformer-rectifier set that is powering the precipitators,

Step 5. Determine effective fly ash resistivity level in the ESP using a correlation that relates fly ash resistivity to ESP current density for each electrical field. Average the results to produce an effective resistivity for the ESP. If this resistivity is not close to, or lower than, the optimum range, proceed with Step 6; otherwise, go to Step 10.

Step 6. Use a correlation relating fly ash composition and flue gas temperature and SO₃ concentration to fly ash resistivity to determine the flue gas SO₃ concentration to needed to produce the optimum fly ash resistivity,

Step 7. Subtract the background SO₃ from the needed SO₃ concentration from Step 6 to determine the amount of SO₃ that must be added to the flue gas to produce the optimum fly ash resistivity.

Step 8. Send rate of additional signal to the controls that operate the SO₃ conditioning system.

Step 9. Repeat Steps 4 and 5.

Step 10. a. If indicated ash resistivity is equal to or less than optimum resistivity, decrease rate of injection by x percent where x is between 5 and 25, or

-   -   b. if indicated ash resistivity is greater than optimum         resistivity, increase rate of injection by x percent where x is         between 5 and 25.

Step 11. Repeat Step 10 until indicated fly ash resistivity passes through optimum resistivity point and then set rate of injection at a point in the range bounded by the levels calculated in the last two interactions, and then

Step 12. Every y minutes, where y is number between 5 and 30, restart the process beginning at Step 2.

DETAILED DESCRIPTION OF THE INVENTION

The herein disclosed invention involves completing a sequence of unique calculations that result in the estimation of the amount of SO₃ that must be added to flue gas to produce optimum fly ash electrical resistivity. This sequence of steps is as follows:

“Typical” Starting Conditions:

a. Low flue gas SO₃ concentration measured at the ESP inlet—0 to 4 ppm. SO₃—example number=3.5.

b. Moderate to high fly ash resistivity—8×10¹⁰ ohm-cm to 5×10¹² ohm-cm.

c. Low ESP power level characterized by low average current densities.

For example, in a three-field electrostatic precipitator the average current densities in the inlet field might be 9.13 na/cm, in the middle field it might be 12.41 na/cm² and in the outlet field, it might be 15.19 na/cm². These current densities correspond to a fly ash resistivity of 1.0×10¹¹ ohm-cm and this level of resistivity is too high to allow optimum ESP performance (see Table 1).

Desired “End” Conditions:

a. Increased flue gas SO₃ measured at ESP inlet—from 2 to 12 ppm, depending on flue gas temperature, flue gas moisture, and fly ash composition.

b. Optimum fly ash resistivity—8×10⁹ ohm-cm to 4×10¹⁰ ohm-cm, depending on ESP collection and reentrainment characteristics—example number 1×10¹⁰ ohm-cm.

c. High ESP power levels as indicated by current density levels.

For example, when the correct level of SO₂ has been added to the flue gas, the average current densities in the ESP would increase to 27.67 na/cm² in the inlet field, 33.50 na/cm² in the middle field and 39.50 na/cm² in the outlet field. The current densities correspond to a fly ash resistivity of 1×10¹⁰ ohm-cm and this level of resistivity should produce optimum ESP performance (see Table 1).

TABLE 1 Typical Per-Field Current Densities for a Range of Resistivies FIRST SECOND THIRD FOURTH FIFTH FIELD 1 FIELD 2 FIELD 3 FIELD 4 FIELD 5 PARAMETER 1 6.255 5.839 5.697 5.018 4.718 PARAMETER 2 0.4813 0.4314 0.4105 0.3405 0.3036 RESISTIVITY CURRENT CURRENT CURRENT CURRENT CURRENT (ohm * cm) na/cm² na/cm² na/cm² na/cm² na/cm² 1.00E+10 27.67 33.50 39.08 41.02 48.08 2.00E+10 19.82 24.84 29.41 32.40 38.96 4.00E+10 14.20 18.42 22.12 25.59 31.57 6.00E+10 11.68 15.46 18.73 22.29 27.91 8.00E+10 10.17 13.66 16.64 20.21 25.58 1.00E+11 9.13 12.41 15.19 18.73 23.90 2.00E+11 6.54 9.20 11.43 14.79 19.36 4.00E+11 4.69 6.82 8.60 11.68 15.69 6.00E+11 3.86 5.73 7.28 10.18 13.87 8.00E+11 3.36 5.06 6.47 9.23 12.71 1.00E+12 3.02 4.59 5.90 8.55 11.88 2.30E+12 2.02 3.21 4.19 6.44 9.23 4.00E+12 1.55 2.53 3.34 5.33 7.80 6.00E+12 1.27 2.12 2.83 4.65 6.90 8.00E+12 1.11 1.87 2.51 4.21 6.32 1.00E+13 1.00 1.70 2.29 3.90 5.90 Note: Resistivities and current densities above the line are in the range that will produce optimum ESP performance. Resistivities and current densities below the line are in the range that will produce suboptimum ESP performance

This invention has several methods to determine the rate of SO₃ addition that will produce the optimum level of fly ash resistivity and hence optimum ESP performance. The first method does not require data feed back from the ESP, while the second method does.

Method 1 is as follows:

Step 1. Obtain the proximate and ultimate analyses of the coal being burned in the boiler and the ash mineral analysis for this coal. Table 2 contains examples of typical analyses.

TABLE 2 Example Coal Composition As Received Example Fly Ash Composition Ultimate Analysis As Constituents (%) (%) Carbon 68.00 LiO2 0.01 Hydrogen 3.86 Na₂O 0.96 Oxygen 6.00 K₂O 2.43 Nitogen 1.00 MgO 0.78 Sulfur 2.20 CaO 2.62 Moisture 3.60 Fe₂O₃ 7.76 Ash 16.34 Al₂O₃ 17.85 SUM 100.00 SiO₂ 61.00 TiO₂ 0.62 P₂O₅ 0.55 SO₃ 2.43 SUM 97.01

Step 2. Determine the average temperature of the flue gas entering the ESP from plant instrumentation. For example, the instrumentation indicates the temperature of the flue gas entering the RSP is 291° F.

Step 3. Estimate SO₃ background level in the flue gas using correlation relating flue gas SO₃ to coal type and coal sulfur content. The SO₃ concentration is calculated as a percentage of SO₂ in the flue gas which can be determined from a combustion calculation using the coal analysis and flue gas O₂ or CO₂ or if the flue gas SO₂ is available from plant instruments, this number can be used in the SO₃ calculation. Using standard, well known chemical formulas and procedures, that calculation is as follows if the assumption for no excess air is used.

A. Calculation of Combustion Products, Air, and O₂ for 100% Combustion.

Required for combustion Ultimate Moles/100 lb fuel Coal analysis Molecular Moles per at 100% total air Constiuent lb/100 lb fuel weight 100 lb fuel Multipliers¹ O₂ Dry Air C 68.00 ÷ 12.01 = 5.662 ×    1.0 and 4.76 5.662 26.951 H₂ 3.86 ÷ 2.02 = 1.911 ×   0.50 and 2.38 0.956 4.548 O₂ 6.00 ÷ 32.00 = 0.188 × −1.00 and −4.76 −0.188 −0.895 N₂ 1.00 ÷ 28.01 = 0.036 S 1.20 ÷ 32.06 = 0.037 ×   1.00 and 4.76 0.037 0.176 H₂O 3.60 ÷ 18.02 = 0.200 Ash 16.34 — — Sum 100.00 8.034 6.467 30.780

A correction for excess air, which is always added to the furnace to ensure complete combustion is next made as follows.

B. Calculation of Air and O₂ for 30% Excess Air (Typical Excess Air Level).

Required for Combustion moles/100 lb fuel at 30% excess air O₂ Dry air O₂ and air × 130/100 total 8.407 40.014 Excess air = 40.014 − 30.780 — 9.234 Excess O₂ = 8.407 − 6.467 1.940 —

Using the values from these two calculations, the final composition of the flue gas is calculated, again using established and well known formulas and procedures.

C. Calculation of Flue Gas Composition.

Products of Combustion Total Flue gas moles/100 % by volume % by volume Constituent Combustion/Fuel/Air lb fuel wet basis dry basis CO₂ 5.662 = 5.662 13.406 14.412 H₂O 1.911 + 0.200 + 0.838^(a) = 2.949 6.983 — SO₂ 0.037 = 0.037 0.088 0.094 N₂ 0.036 + 31.611^(b) = 31.647 74.931 80.555 O₂ 1.940 = 1.940 4.593 4.938 Sum wet 42.235 Sum dry = 42.235 − 2.949 39.286 ^(a)Moles H₂O in air = (40.014 × 29 × 0.013) ÷ 18 = 0.838 ^(b)Moles N₂ in air = (40.014 × 0.79) = 31.611

The critical numbers from these calculations are the SO₂ concentrations:

-   -   0.088%, wet basis, and     -   094%, dry basis.         The moisture concentration, 6.98% is also critical Once these         numbers are known, the native SO₃ concentration in the flue gas         can be calculated as follows:

The SO₂ concentration dry (the resistivity concentration in this example uses the equivalent SO₃ concentration of “dry” flue gas) is equal to 0.094%. the appropriate SO₂ to SO₃ conversion factor for this coal is 0.4% so the approximate SO₃ concentration is: 0.00094×0.004=3.76 PPM (dry basis) As an alternative, the flue gas SO₂ concentration can be obtained from the plant's Continuous Emissions Monitoring (CEM) system, corrected for flue gas moisture concentration using factors from the combustion calculation and multiplied by the factor 0.004 to estimate to inherent or background SO₃ concentration. For other coals, for example, western coals, the appropriate conversion factor is 0.001 and for Powder River Basin Coals, the conversion factor is 0.005 (as opposed to 0.004).

Step 4. Calculate the base ash resistivity using empirical equations relating ash resistivity to ash composition, flue gas moisture and flue gas temperature. The Bickelhaupt equations are an example of relationships that can be used for this calculation. This particular calculation is made using the ash mineral analysis from Table 2 and the moisture and SO₃ calculations from step 2 using the following sequence of substeps:

-   Substep 1: Normalize the weight percentages to sum 100% by dividing     each specified percentage by the sum of the specified percentages. -   Substep 2: Divide each oxide percentage by the respective molecular     weight to obtain the mole fractions. -   Substep 3: Divide each mole fraction by the sum of the mole     fractions and multiply by 100 to obtain the molecular percentages as     oxides. -   Substep 4: Multiply each molecular percentage by the decimal     fraction of cations in the given oxide to obtain the atomic     concentrations.     These substeps are illustrated for the example ash in the following     table.

Atomic Specified Normalized Molecular Mole Molecular Cationic Concentration Oxide Weight % Weight % Weight Fraction Percentage Fraction Of Cation Li₂O 0.01 0.01 29.88 0.00034 0.024 0.67 0.016 Na₂O 0.96 0.99 61.98 0.01600 1.116 0.67 0.744 K₂O 2.43 2.50 94.20 0.02654 1.854 0.67 1.236 MgO 0.78 0.80 40.31 0.01985 1.387 0.50 0.694 CaO 2.62 2.70 56.08 0.04815 3.364 0.50 1.682 Fe₂O₃ 7.76 8.00 159.70 0.05009 3.500 0.40 1.400 Al₂O₃ 17.85 18.40 101.96 0.18046 12.608 0.40 5.043 SiO₂ 61.00 62.89 60.09 1.04660 73.123 0.33 24.368 TiO₂ 0.62 0.64 79.90 0.00801 0.560 0.33 0.186 P₂O₅ 0.55 0.57 141.94 0.00402 0.281 0.29 0.080 SO₃ 2.43 2.50 80.06 0.03123 2.183 0.25 0.546 Sum 97.01 100.00 1.43129 100.000

Now that the atomic concentrations of the critical ash mineral constituents are known, the rest of the calculation proceeds by calculating three separate resistivities, the volume resistivity, ρ_(v), the surface resistivity, ρ_(s), and the acid resistivity, ρ_(a). These three resistivities are then combined to give the net resistivity of the ash using the parallel resistance formula. For the example coal, the calculation proceeds as follows using the following formulas and definitives.

Bickelhaupt Equations

ρ_(v)=exp[−1.8916 ln X−0.9696 ln Y+1.234 ln Z+3.62876−(0.069078)E+9980.58/T] ρ_(s)=exp[27.59774−2.233348 ln X−(0.00176)W−(0.069078)E−(0.00073895)(W)exp(2303.3/T)] ρ_(a)=exp[85.1405−(0.708046)CSO₃−23267.2/T−(0.069078)E], for z<3.5% or K>1.0% ρ_(a)=exp[59.0677−(0.854721)CSO₃−13049.47/T−(0.069078)E], for z<3.5% or K>1.0% 1/ρ_(vs)=1/ρ_(v)+1/ρ_(s) 1/ρ_(vsa)=1/ρ_(vs)+1/ρ_(a)

ρ_(v)=volume resistivity (ohm-cm)

ρ_(s)=surface resistivity (ohm-cm)

ρ_(a)=adsorbed acid resistivity (ohm-cm)

ρ_(vs)=volume and surface resistivity (ohm-cm)

ρ_(vsa)=total resistivity (ohm-cm)

X=Li+Na percent atomic concentration

Y=Fe percent atomic concentration

Z=Mg+Ca percent atomic concentration

K=K percent atomic concentration

T=absolute temperature (K)

W=moisture in flue gas (volume %)

CSO₃=concentration of SO₃ (ppm, dry)

E=applied electric field (kV/cm)

Using the above definitions, equations and calculated values, the calculation proceeds for the example case as follows:

X=0.016+0.744=0.76

Y=1.40

Z=0.694+1.682=2.376

K=1.236

T=417 (Example gas temperature 291° F.)

W=6.983

CSO₃=(from Calculation 2)3.76 ppm, dry

E=10(typical electric field value) ρ_(v)=exp[−1.8916 ln(0.76)−0.9696 ln(1.40)+1.237 ln(2.376)+3.62876−(0.069078)(10)+9980.58/417] =1.636×10¹² ohm-cm ρ_(s)=exp[27.59774−2.23348 ln(0.76)−(0.00176)(6.983)−(0.069078)(10)−(0.00073895)(6.983)exp(2303.0/417)] =2.392×10¹¹ ohm-cm ρ_(a)=exp[85.1405−(0.708046)(3.76)−23267.2/417−(0.069078)(10)] =1.939×10¹¹ ohm-cm 1/ρ_(vs)=1/1.636×10¹²+1/2.392×10¹¹=4.792×10⁻¹² ρ_(vs)=2.1×10¹¹ ohm-cm 1/ρ_(vsa)=1/4.792×10⁻¹²+1/1.939×10¹¹=9.949×10⁻¹² ρ_(vsa)=1.0 ×10¹¹ ohm-cm

In this example, the calculated resistivity is found to be 1.0×10¹¹ ohm-cm, which is too high for optimum ESP performance, so additional SO₃ must be added to the flue gas.

Step 5. Use a correlation relating the base fly ash resistivity and flue gas SO₃ concentration to determine the flue gas SO₃ concentration needed to produce the optimum fly ash resistivity.

From the preceding step, the relationships between the acid resistivity, surface resistivity, volume resistivity and net ash resistivity are known. Further, it is known that the desirable level of resistivity is 1.0×10¹⁰ ohm-cm. Hence, the calculation proceeds as follows: 1/ρ_(vsa)=1/ρ_(vs)+1/ρ_(a) where ρ_(vsa)=1×10¹⁰ ohm-cm(the desirable ρ) ρ_(vs)=2.1×10¹¹ ohm-cm(from preceding calculation)

$\begin{matrix} {{1/\rho_{a}} = {{1/\rho_{vsa}} - {1/\rho_{vs}}}} \\ {= {{1.0 \times 10^{10}} - {2.761905 \times 10^{12}}}} \\ {= {9.5238 \times 10^{11}}} \end{matrix}$ ρ_(va)=1.05×10¹⁰ ohm-cm

also from the preceding calculation, ρ_(a)=exp[85.1405−(0.708046)CSO₃−23267.2/T−(0.069078)E] where T=417 (from preceding calculation)

E=10 (from preceding calculation)

hence 1.05×10¹⁰=exp[85.1405−(0.708046)CSO₃−23267.2/417−(0.069078)(10)] ln(1.05×10¹⁰)=85.1405−(0.708046)CSO₃−55.79664−0.69078 23.07464109=28.652−(0.708046)CSO₃ (0.708046)CSO₃=5.578 CSO₃=7.878 Correcting for wet conditions SO₃ needed=7.878×(39.286/42.235)=7.33 ppm

Step 6. Subtract the background SO₃ concentration from the needed SO₃ concentration from the needed SO₃ that must be added to the flue gas to produce the optimum fly ash resistivity.

The combustion calculation results and the background SO₃ calculation in step 2, the SO₃ concentration is estimated to be 0.00088×0.004=3.52 ppm (wet basis). From the desired level calculation above, the desirable SO₃ level=7.33 ppm, hence the difference=7.33−3.52=3.81 ppm. Hence, 3.81 ppm, SO₃ must be added to the flue gas to produce the desired level of fly ash resistivity.

Step 7. Send rate of addition signal to the controls that operate the SO₃ conditioning system. In this case, the signal should be sent that will cause the SO₃ conditioning system to add 3.8 ppm SO₃ to the flue gas.

Notice that this procedure uses the equations developed by Dr. Bickelhaupt to relate flue gas composition and ash mineral analysis in the calculations, but any set equations relating flue gas SO₃ concentrations and ash mineral analysis to fly ash resistivity could be used. For example, the equations developed by Joe McCain and published in EPRI technical report 1004075 can be used.

This concludes Method 1.

Method 2 is as follows:

The example calculation for this method resumes the same starting conditions that were assumed for method 1. They are as follows:

a. Low flue gas SO₃ concentration measured at the ESP inlet—0 to 4 ppm: SO₃ —example number=3.5.

b. Moderate to high fly ash resistivity—8×10¹⁰ ohm-cm to 5×10¹² ohm-cm.

c. Low ESP power level characterized by low average current densities.

For example, in a three-field electrostatic precipitator the average current densities in the inlet field might be 9.13 na/cm², in the middle field it might be 12.41 na/cm² and the outlet field might be 15.19 na/cm². These current densities correspond to a fly ash resistivity of 1.0×10¹¹ ohm-cm and this level of resistivity is too high to allow optimum ESP performance (see Table 1).

Typical Per-Field Current Densities for a Range of Resistivies FIRST SECOND THIRD FOURTH FIFTH FIELD 1 FIELD 2 FIELD 3 FIELD 4 FIELD 5 PARAMETER 1 6.255 5.839 5.697 5.018 4.718 PARAMETER 2 0.4813 0.4314 0.4105 0.3405 0.3036 RESISTIVITY CURRENT CURRENT CURRENT CURRENT CURRENT (ohm * cm) na/cm² na/cm² na/cm² na/cm² na/cm² 1.00E+10 27.67 33.50 39.08 41.02 48.08 2.00E+10 19.82 24.84 29.41 32.40 38.96 4.00E+10 14.20 18.42 22.12 25.59 31.57 6.00E+10 11.68 15.46 18.73 22.29 27.91 8.00E+10 10.17 13.66 16.64 20.21 25.58 1.00E+11 9.13 12.41 15.19 18.73 23.90 2.00E+11 6.54 9.20 11.43 14.79 19.36 4.00E+11 4.69 6.82 8.60 11.68 15.69 6.00E+11 3.86 5.73 7.28 10.18 13.87 8.00E+11 3.36 5.06 6.47 9.23 12.71 1.00E+12 3.02 4.59 5.90 8.55 11.88 2.30E+12 2.02 3.21 4.19 6.44 9.23 4.00E+12 1.55 2.53 3.34 5.33 7.80 6.00E+12 1.27 2.12 2.83 4.65 6.90 8.00E+12 1.11 1.87 2.51 4.21 6.32 1.00E+13 1.00 1.70 2.29 3.90 5.90 Note: Resistivities and current densities above the line are in the range that will produce optimum ESP performance. Resistivities and current densities below the line are in the range that will produce suboptimum ESP performance

As in the Method 1 example calculation, the desired end point is the same. It is described in the following paragraph:

Desired “End” Conditions:

a. Increased flue gas SO₃ measured at ESP inlet—from 2 to 12 ppm, depending on flue gas temperature, flue gas moisture, and fly ash composition.

b. Optimum fly ash resistivity—8×10⁹ ohm-cm to 4×10¹⁰ ohm-cm, depending on ESP collection and reentrainment characteristics—example number 1×10¹⁰ ohm-cm.

c. High ESP power levels as indicated by current density levels.

For example, when the correct level of SO₃ has been added to the flue gas, the average current densities in the ESP would increase to 27.67 nA/cm² in the inlet field, 33.50 na/cm² in the middle field, and 39.08 na/cm² in the outlet field. The current densities correspond to a fly ash resistivity of 1×10¹⁰ ohm-cm and this level of resistivity should produce optimum ESP performance (see Table 1).

Method 2 uses the following alternative sequence of steps to determine the optimum injection rate for SO₃:

Step 1. Obtain the proximate and ultimate analysis of the coal being burned in the boiler and the ash mineral analysis for this coal. Table 2 contains examples of typical analysis.

TABLE 2 Example Coal Composition As Received Example Fly Ash Composition Ultimate Analysis As Constituents (%) (%) Carbon 68.00 LiO2 0.01 Hydrogen 3.86 Na₂O 0.96 Oxygen 6.00 K₂O 2.43 Nitogen 1.00 MgO 0.78 Sulfur 2.20 CaO 2.62 Moisture 3.60 Fe₂O₃ 7.76 Ash 16.34 Al₂O₃ 17.85 SUM 100.00 SiO₂ 61.00 TiO₂ 0.62 P₂O₅ 0.55 SO₃ 2.43 SUM 97.01

Step 2. Determine the average temperature of the flue gas entering the ESP from plant instrumentation. For example, the instrumentation indicates the temperature of the flue gas entering the ESP is 291° F.

Step 3. Estimate SO₃ background level in the flue gas using correlation relating flue gas SO₃ to coal type and coal sulfur content. The SO₃ concentration is calculated as a percentage of SO₂ in flue gas which can be determined from a combustion calculation using the coal analysis and flue gas O₂ or CO₂ or if the flue gas SO₂ is available from plant instruments, this number can be used in the SO₃ calculation. Using standard, well known chemical formulas and procedures, that calculation is as follows if the assumption for no excess air is used.

A. Calculation of Combustion Products, Air, and O₂ for 100% Combustion.

Required for combustion Ultimate Moles/100 lb fuel Coal analysis Molecular Moles per at 100% total air Constiuent lb/100 lb fuel weight 100 lb fuel Multipliers¹ O₂ Dry Air C 68.00 ÷ 12.01 = 5.662 ×    1.0 and 4.76 5.662 26.951 H₂ 3.86 ÷ 2.02 = 1.911 ×   0.50 and 2.38 0.956 4.548 O₂ 6.00 ÷ 32.00 = 0.188 × −1.00 and −4.76 −0.188 −0.895 N₂ 1.00 ÷ 28.01 = 0.036 S 1.20 ÷ 32.06 = 0.037 ×   1.00 and 4.76 0.037 0.176 H₂O 3.60 ÷ 18.02 = 0.200 Ash 16.34 — — Sum 100.00 8.034 6.467 30.780

A correction for excess air, which is always added to the furnace to ensure complete combustion is next made as follows:

B. Calculation of Air and O₂ for 30% Excess Air (Typical Excess Air Level).

Required for Combustion moles/100 lb fuel at 30% excess air O₂ Dry air O₂ and air × 130/100 total 8.407 40.014 Excess air = 40.014 − 30.780 — 9.234 Excess O₂ = 8.407 − 6.467 1.940 —

Using the values from these two calculations, the final composition of the flue gas is calculated, again using established and well known formulas and procedures.

C. Calculation of Flue Gas Composition.

Products of Combustion Total Flue gas moles/100 % by volume % by volume Constituent Combustion/Fuel/Air lb fuel wet basis dry basis CO₂ 5.662 = 5.662 13.406 14.412 H₂O 1.911 + 0.200 + 0.838^(a) = 2.949 6.983 — SO₂ 0.037 = 0.037 0.088 0.094 N₂ 0.036 + 31.611^(b) = 31.647 74.931 80.555 O₂ 1.940 = 1.940 4.593 4.938 Sum wet 42.235 Sum dry = 42.235 − 2.949 39.286 ^(a)Moles H₂O in air = (40.014 × 29 × 0.013) ÷ 18 = 0.838 ^(b)Moles N₂ in air = (40.014 × 0.79) = 31.611

The critical numbers from these calculations are the SO₂ concentrations:

-   -   0.0870, wet basis, and     -   0.0970 dry basis.

The moisture concentration 6.970 is also critical.

Once these numbers are known, the native SO₃ concentration in the flue gas can be calculated as follows:

The SO₂ concentration dry (the resistivity concentration in this example uses the equivalent SO₃ concentration “dry” flue gas) is equal to 0.094%. The appropriate SO₂ to SO₃ conversion factor for this coal is 0.4% so the approximate SO₃ concentration is: 0.00094×0.004=3.76 PPM (dry basis). As an alternative, the flue gas SO₂ concentration can be obtained from the plant's Continuous Emissions Monitoring (CEM) system, corrected for flue gas moisture concentration using factors from the combustion calculation and multiplied by the factor 0.004 to estimate inherent or background SO₃ concentration. For other coals, for example, western coals, the approximate conversion factor is 0.001 and for Powder River Basin Coals, the conversion factor is 0.005 (as apposed to 0.004).

To this point, the calculations for Method 1 and Method 2 are the same, however, they are different from this point on.

Step 4. The secondary current applied to the electrostatic precipitator is obtained from the controls for each transformer-rectifier set that is powering the precipitator. These current numbers are translated into current densities by dividing the plate area powered by the transformer-rectifier set. In this example-case, the precipitator has four electrical fields in the direction of gas flow with four transformer-rectifier sets per field.

Readings from the transformer/rectifier sets are as follows:

TR1 TR2 TR3 TR4 Field 1 165 ma 165 ma 165 ma 165 ma Field 2 224 ma 224 ma 224 ma 224 ma Field 3 274 ma 274 ma 274 ma 274 ma Field 4 338 ma 338 ma 338 ma 338 ma In this example-case, each transformer/rectifier set energized 19,440 ft² of plate area. For a typical field, these currents translate into current densities as follows: 165 ma×(1.0×10⁻³ ma/a)/(19,440 ft²)=8.488×10⁻⁶ a/ft²=8.488 μa/ft² 8.488×10⁻⁶×1.076=9.133 na/cm² Note: 1.0 μa/ft²=1.076 na/cm² Similar calculations can be used to produce the following table

TR1 TR2 TR3 TR4 Field 1 9.13 9.13 9.13 9.13 Field 2 12.41 12.41 12.41 12.41 Field 3 15.19 15.19 15.19 15.19 Field 4 18.73 18.73 18.73 18.73 Where the units are nA/cm²

Notice that in this example, all of the TR sets in the same field have been assumed to have the same operating point, i.e., the same voltage and current levels. If these numbers were different, an averaging process, in Step 5, would be used to deal with this more common situation.

Step 5. Determine effective fly ash resistivity level in the ESP using a correlation that relates fly ash resistivity to ESP current density for each electrical field. Average the results to produce an effective resistivity for the ESP. If this resistivity is close to, or lower than, the optimum range, go to Step 10, otherwise proceed to Step 6.

In this example-case, the correlations published in EPRI report CS-5040, table 3–4 are used.

These correlations, after amplification are as follows:

Field 1 log₁₀ (J, nA/cm²) = (6.455 ± 0.370) − 0.5013 log₁₀ (ρ, ohm-cm) Field 2 log₁₀ (J, nA/cm²) = (6.839 ± 0.360) − 0.5214 log₁₀ (ρ, ohm-cm) Field 3 log₁₀ (J, nA/cm²) = (5.497 ± 0.304) − 0.3905 log₁₀ (ρ, ohm-cm) Field 4 log₁₀ (J, nA/cm²) = (5.718 ± 0.327) − 0.4005 log₁₀ (ρ, ohm-cm) Field 5 log₁₀ (J, nA/cm²) = (3.328 ± 0.306) − 0.1736 log₁₀ (ρ, ohm-cm) Where J is in nA/cm² and ρ is in ohm-cm. Since, log(e)=1/ln(10) substitution gives: log(J)=log(e)ln(J)=ln(J)/ln(10)=ln(J)/2.302585 similarly, log(ρ)=ln(ρ)/ln(10)=ln(ρ)/2.302585 and further substitution gives:

Field 1 ln(J)/ln(10) = 6.455 − 0.5013 ln(ρ)/ln(10) or ln(J) = 2.302585 × 6.455 − 0.5013 ln(ρ) Field 1 ln(J) = 14.8632 − 0.5013 ln(ρ) similarly Field 2 ln(J) = 15.74738 − 0.5214 ln(ρ) Field 3 ln(J) = 12.65731 − 0.3905 ln(ρ) Field 4 ln(J) = 13.16618 − 0.4005 ln(ρ) Field 5 ln(J) = 7.66300 − 0.1736 ln(ρ) These equations are inverted to give the following:

Field 1 ln(ρ) = 29.64931 − 1.994813 ln(J) Field 2 ln(ρ) = 30.20211 − 1.917913 ln(J) Field 3 ln(ρ) = 32.41309 − 2.560819 ln(J) Field 4 ln(ρ) = 32.87435 − 2.496879 ln(J) Field 5 ln(ρ) = 44.14171 − 5.76037 ln(J) From Calculation 3, we have the following:

J Field 1  9.13 na/cm² Field 2 12.41 na/cm² Field 3 15.19 na/cm² Field 4 18.73 na/cm² Using the ρ vs. J equation gives:

ρ Field 1  9.1 × 10¹⁰ ohm-cm Field 2 10.4 × 10¹⁰ ohm-cm Field 3 11.3 × 10¹⁰ ohm-cm Field 4 16.2 × 10¹⁰ ohm-cm Average 11.8 × 10¹⁰ ohm-cm Note that the resistivity is much higher than the optimum value of 10¹⁰ ohm-cm.

Step 6. Use a correlation relating fly ash composition and flue gas temperature and SO₃ concentration to fly ash resistivity to determine the flue gas SO₃ concentration to needed to produce the optimum fly ash resistivity.

That calculation proceeds in a sequence of substeps as follows using the equations developed by Dr. Bickelhaupt and published in EPRI report C9-4145, Appendix A. Starting with the example ash composition in Table 2, complete substep as follows:

-   Substep 1: Normalize the weight percentages to sum 100% by dividing     each specified percentage by the sum of the specified percentages. -   Substep 2: Divide each oxide percentage by the respective molecular     weight to obtain the mole fractions. -   Substep 3: Divide each mole fraction by the sum of the mole     fractions and multiply by 100 to obtain the molecular percentages as     oxides. -   Substep 4: Multiply each molecular percentage by the decimal     fraction of cations in the given oxide to obtain the atomic     concentrations.     All of these sub-steps are illustrated in the following table for     the data in Table 2.

Atomic Specified Normalized Molecular Mole Molecular Cationic Concentration Oxide Weight % Weight % Weight Fraction Percentage Fraction Of Cation Li₂O 0.01 0.01 29.88 0.00034 0.024 0.67 0.016 Na₂O 0.96 0.99 61.98 0.01600 1.116 0.67 0.744 K₂O 2.43 2.50 94.20 0.02654 1.854 0.67 1.236 MgO 0.78 0.80 40.31 0.01985 1.387 0.50 0.694 CaO 2.62 2.70 56.08 0.04815 3.364 0.50 1.682 Fe₂O₃ 7.76 8.00 159.70 0.05009 3.500 0.40 1.400 Al₂O₃ 17.85 18.40 101.96 0.18046 12.608 0.40 5.043 SiO₂ 61.00 62.89 60.09 1.04660 73.123 0.33 24.368 TiO₂ 0.62 0.64 79.90 0.00801 0.560 0.33 0.186 P₂O₅ 0.55 0.57 141.94 0.00402 0.281 0.29 0.080 SO₃ 2.43 2.50 80.06 0.03123 2.183 0.25 0.546 Sum 97.01 100.00 1.43129 100.000 Using the % atomic concentrations from the above calculations, use the following equations for calculation of fly ash resistivity (Bickelhaupt equations). ρ_(v)=exp[−1.8916 ln X−0.9696 ln Y+1.234 ln Z+3.62876−(0.069078)E+9980.58/T] ρ_(s)exp[27.59774−2.233348 ln X−(0.00176)W−(0.069078)E−(0.00073895)(W)exp(2303.3/T)] ρ_(a)exp[85.1405−(0.708046)CSO₃−23267.2/T−(0.069078)E], for z<3.5% or K>1.0% ρ_(a)=exp[59.0677−(0.854721)CSO₃−13049.47/T−(0.069078)E], for z>3.5% and K<1.0% 1/ρ_(vs)=1/ρ_(v)+1/ρ_(s) 1/ρ_(vsa)=1/ρ_(vs)+1/ρ_(a)

ρ_(v)=volume resistivity (ohm-cm)

ρ_(s)=surface resistivity (ohm-cm)

ρ_(a)=adsorbed acid resistivity (ohm-cm)

ρ_(vs)=volume and surface resistivity (ohm-cm)

ρ_(vsa)=total resistivity (ohm-cm)

X=Li+Na percent atomic concentration

Y=Fe percent atomic concentration

Z=Mg+Ca percent atomic concentration

K=K percent atomic concentration

T=absolute temperature (K)

W=moisture in flue gas (volume %)

CSO₃=concentration of SO₃ (ppm, dry)

E=applied electric field (kV/cm)

For the example case,

X=0.016+0.744=0.76

Y=1.40

Z=0.694+1.682=2.376

K=1.236

T=417 (Example gas temperature 291° F.)

W=6.983

CSO₃=(from Calculation 2)3.76 ppm, dry

E=10(typical electric field value) ρ_(v)=exp[−1.8916 ln(0.76)−0.9696 ln(1.40)+1.237 ln(2.376)+3.62876−(0.069078)(10)+9980.58/417] =1.636×10¹² ohm-cm ρ_(s)=exp[27.59774−2.23348 ln(0.76)−(0.00176)(6.983)−(0.069078)(10)−(0.00073895)(6.983)exp(2303.0/417)] =2.392×10¹¹ ohm-cm

$\begin{matrix} {\rho_{a} = {\exp\mspace{11mu}\left\lbrack {85.1405 - {(0.708046)(3.76)} - {23267.2/417} - {(0.069078)(10)}} \right\rbrack}} \\ {= {{1.939 \times 10^{11}{ohm}} - {cm}}} \\ \; \end{matrix}$ 1/ρ_(vs)=1/1.636×10¹²+1/2.392×10¹¹=4.792×10⁻¹² ρ_(vs)=2.1 ×10¹¹ ohm-cm 1/ρ_(vsa)=1/4.792×10⁻¹²+1/1.939×10¹¹=9.949×10⁻¹² ρ_(vsa)=1.0 ×10¹¹ ohm-cm

This resistivity is consistent with the resistivity calculated from the precipitator current densities, but this consistency is not required for Method 2 since this calculation is being used to obtain an approximate SO₃ injection rate which will be refined in the following steps. The approximate level of SO₃ injection is calculated as follows:

From the preceding calculation, 1/ρ_(vsa)=1/ρ_(vs)+1/ρ_(a) where ρ_(vsa)=1×10¹⁰ ohm-cm(the desirable ρ) ρ_(vs)=2.1×10¹¹ ohm-cm(from preceding calculation)

$\begin{matrix} {{1/\rho_{a}} = {{1/\rho_{vsa}} - {1/\rho_{vs}}}} \\ {= {{1.0 \times 10^{10}} - {2.761905\; \times 10^{12}}}} \\ {= {9.5238\; \times 10^{11}}} \end{matrix}$ ρ_(va)=1.05×10¹⁰ ohm-cm

also from Calculation 6, ρ_(a)=exp[85.1405−(0.708046)CSO₃−23267.2/T−(0.069078)E] where T=417 (from preceding calculation)

E=10 (from preceding calculation)

hence 1.06×10¹⁰=exp[85.1405−(0.708046)CSO₃−23267.2/417−(0.069078)(10)] ln(1.05×10¹⁰)=85.1405−(0.708046)CSO₃−55.79664−0.69078 23.07464109=28.652−(0.708046)CSO₃ (0.708046)CSO₃=5.578 CSO₃=7.878 Correcting for wet conditions Hence, the approximate total SO₃ needed=7.878×(39.286/42.235) =7.33 ppm

Step 7. Subtract the background SO₃ from the needed SO₃ concentration from Step 6 to determine the amount of SO₃ that must be added to the flue gas to produce the optimum fly ash resistivity. That calculation, for the example-case, proceeds as follows:

The SO₃ from combustion calculation and background calculation, =0.00088×0.004=3.52 ppm (wet basis)

From the calculation above, the approximate desirable SO₃ level=7.33 ppm. Difference=7.33−3.52=3.81 ppm.

This calculation shows that approximately 3.8 ppm of SO₃ should be added to the flue gas to produce an optimum level of fly ash resistivity.

Consequently, output to SO₃ control system a signal that will raise the SO₃ level in the flue gas by 3.8 ppm.

Step 8. Send rate of additional signal to the controls that operate the SO₃ conditioning system.

Step 9. Repeat Steps 4 and 5.

Step 10.

a. If indicated ash resistivity is equal to or less than optimum resistivity, decrease rate of injection by x percent where x is between 5 and 25.

Or

b. If indicated ash resistivity is greater than optimum resistivity, increase rate of injection by x percent where x is between 5 and 25.

Step 11. Repeat Step 10 until indicated fly ash resistivity passes through optimum resistivity point and then set rate of injection at a point in the range bounded by the levels calculated in the last two interactions; for example, at a point that is halfway between the two levels.

Step 12. Every y minutes, where y is number between 5 and 30, restart the process beginning at Step 2.

Obviously, many modifications may be made without departing from the basic spirit of the present invention. 

1. A method for determining a most effective injection rate for SO₃ into a flue gas resulting from a coal burned in a boiler, the method comprising: 1, obtaining proximate and ultimate analyses of the coal being burned in the boiler and ash mineral analysis for the coal, 2, determining an average temperature of the flue gas entering an electrostatic precipitator (ESP) from plant instrumentation, wherein the ESP is powered with multiple electrical fields, 3, estimating SO₃ background level in the flue gas using correlation relating flue gas SO₃ to coal type and coal sulfur content, 4, obtaining a value of a current applied to the ESP from controls for each transformer-rectifier set that is powering the precipitator, 5, determining an effective fly ash resistivity level in the ESP using a correlation that relates fly ash resistivity to ESP current density for each electrical field, and averaging the results to produce the effective resistivity of the fly ash collected in the ESP, 6, a, if the indicated ash resistivity is equal to or less than an optimum resistivity, decreasing the rate of injection by x percent where x is between 5 and 25, or b, if the indicated ash resistivity is greater than the optimum resistivity, increasing the rate of injection by x percent where x is between 5 and 25, 7, repeating step 6 until the indicated fly ash resistivity passes through an optimum resistivity point, setting the rate of injection at a point in the range bounded by the levels calculated in the last two interactions, and then 8, every y minutes, where y is number between 5 and 30, restarting the process beginning at step
 2. 2. A method for determining a most effective injection rate for SO₃ into a flue gas resulting from a coal burned in a boiler, the method comprising: 1, obtaining proximate and ultimate analyses of the coal being burned in the boiler and ash mineral analysis for the coal, 2, determining an average temperature of the flue gas entering an electrostatic precipitator (ESP) from plant instrumentation, wherein the ESP is powered with multiple electrical fields, 3, estimating SO₃ background level in the flue gas using correlation relating flue gas SO₃ to coal type and coal sulfur content, 4, obtaining a value of a current applied to the ESP from controls for each transformer-rectifier set that is powering the precipitator, 5, determining an effective fly ash resistivity level in the ESP using a correlation that relates fly ash resistivity to ESP current density for each electrical field, averaging the results to produce the effective resistivity of the fly ash collected in the ESP, and if this resistivity is not close to, or lower than, the optimum range, proceeding with step 6; otherwise, going to step 10, 6, using a correlation relating fly ash composition and flue gas temperature and SO₃ concentration to the fly ash resistivity to determine a flue gas SO₃ concentration needed to produce the optimum fly ash resistivity, 7, subtracting the background SO₃ from the needed SO₃ concentration from step 6 to determine the amount of SO₃ that must be added to the flue gas to produce the optimum fly ash resistivity, 8, sending a signal to the controls that operate an SO₃ conditioning system to increase the amount of SO₃ injected into the flue gas by the additional amount determined in step 7, 9, repeating steps 4 and 5, 10, a, if the indicated ash resistivity is equal to or less than the optimum resistivity, decreasing the rate of injection by x percent where x is between 5 and 25, or b, if the indicated ash resistivity is greater than the optimum resistivity, increasing the rate of injection by x percent where x is between 5 and 25, 11, repeating step 10 until the indicated fly ash resistivity passes through the optimum resistivity point, setting the rate of injection at a point in the range bounded by the levels calculated in the last two interactions, and then 12, every y minutes, where y is number between 5 and 30, restarting the process beginning at step
 2. 